I have this transfer function:

I'm not sure how to get the +5 on the top, I know of some circuits where i can get everything but that but I dont know where to go from there

 

Are you being asked to determine the circuit that would have resulted in the given transfer equation?

If so, that is an interesting twist.

hgmjr

 

Yes the question states: Using op amps, resistors, and capacitors design a circuit that has a transfer function of....

 

I have this transfer function:

I'm not sure how to get the +5 on the top, I know of some circuits where i can get everything but that but I dont know where to go from there

This can be implemented by concatenating a bandpass filter with a PI-filter (proportional-integral, not the Greek letter pi).

Please see the following reference for the bandpass circuit. I'm almost positive that you know about the bandpass, but perhaps the missing piece for you is the PI-filter which is less well-known.

http://www.physics.uq.edu.au/people/jones/phys2810/ph248tutes/circuits/sol5/node2.html

A PI-filter can be made from a standard inverting op-amp amplifier which normally has transfer function -R2/R1. Simply replace R2 with a series combination of R2 and a capacitor C.

The transfer function of the PI-filter will be \( T(s)=\Bigl( {{-R_2}\over{R_1}} \Bigr) {{s+{{1}\over{R_2 C}}}\over{s}}\)

Note that the s in the denominator of the PI-filter will cancel with the s in the numerator of the bandpass filter. You'll have to play around with the gains and the poles and zero locations to make it match. Also, you may need an inverter stage depending on the exact circuits you choose.

 

so take these two and sum them together?

 

so take these two and sum them together?

Sorry, I should have been more clear. Concatenate means to put the two stages in series, which will multiply their individual transfer functions. In other words the input signal might go into the PI-filter first, and then the output of the PI-filter would go into the input of the bandpass filter. The output of the bandpass filter would then be the output of the entire system.

Alternatively, you could place the bandpass filter before the PI-filter. In this case the output of the bandpass filter will go into the input of the PI-filter. The input to the bandpass filter would then be the input of the system and the output of the PI-filter would be the output of the system.

 

I dont see how multiplying them together will give me the constant i need on top. The circuit from the link you gave me has an s on the top so multiplying that into something will just give me that constant time s.

OR will i be able to divide an s out of each term?

 

Another "simplistic" approach is to form the State-Space representation of the transfer function.

Using integrators [op amps with capacitive feedback and resistive input], feedback summing junctions & gain stages one can implement any arbitrary transfer function. Clearly it doesn't give the most concise circuit implementation but higher order TF polynomial implementations (3rd, 4th etc) are no more difficult than lower order cases - in principle.

 

OR will i be able to divide an s out of each term?

Yes, that's exactly it. The pole and the zero at s=0 will cancel out. You will be left with exactly what you want.

By the way t_n_k's method is a very good idea. However, I don't know if you have studied the state space representation of systems yet.

As he said, it is not the most concise approach, but neither is the one I suggested.

 

thanks for the help guys, I went steveb's route and with the transfer function i came up with for the circuit it was impossible to get the values I needed but i'm just going to leave it as is